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Let $X_1 , ... , X_n$ be a series of independent random variables following a Bernoulli distribution with parameter $\theta$. And let $S_n = \sum_1^n X_i$.
We know an unbiased estimator of the variance for the Bernoulli distribution: $$1/2 * (X_1 - X_2)^2$$

Using the Rao-Blackwell Theorem find the best estimator: (Improved from Rao BlackWell)

$$Z = E_\theta(1/2*(X_1 - X_2)^2 | S_n)$$

Sorry for the poor translation of the problem, the original text is in french. Does anyone have any idea of how to do that? So far neither my book or internet haven't been really helpful... Thanks in advance

Community
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clds
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  • Do you know the Lehmann-Scheffe theorem? – Potato Dec 25 '14 at 13:15
  • Not yet! Back to studies! – clds Dec 25 '14 at 13:35
  • My answer below sketches how I would do it. Perhaps there is a simpler way I am missing. – Potato Dec 25 '14 at 13:37
  • I rolled back the edit because it gave the wrong expression for the estimator. – Potato Dec 25 '14 at 13:37
  • The question seems to ask to compute $Z=E_\theta(\frac12(X_1-X_2)^2|S_n)$ but you accepted an answer which does not do that. Please explain. – Did Dec 27 '14 at 02:45
  • Sorry, I'm new here. The answer did helped me to get to the result and I then posted my calculus latter as a separate answer, should I set it as the correct answer? Also I can't accept my own answer for the next 2 hours – clds Dec 27 '14 at 10:15

2 Answers2

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It suffices to find an unbiased estimator that is a function of a complete, sufficient statistic. (This is the Lehmann-Scheffe theorem). The Rao-Blackwell theorem can often be used to guess such estimators. You take some known estimator and condition it on sufficient statistic to give you a "better" estimator that is also a function of the sufficient statistic. Many times, but not always, the new estimator is the UMVUE ("best estimator"). I believe this is the case here, though I have not checked the details. To complete the proof, it suffices to check that the new estimator is unbiased (so that Lehmann-Scheffe applies).

Potato
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Using Rao-Blackwell:

$Z = 1/2 * (E_\theta((X_1 - X_2)^2 | S_n))$
$Z = 1/2 * (E(X_1^2 | S_n) + E(X_2^2 | S_n) - 2 * E(X_1 * X_2 | S_n))$

Considering that $X_1 {\buildrel d \over =} X_i$ we can then write:

$Z = 1/2 * (E(X_i^2 | S_n) + E(X_j^2 | S_n) - 2 * E(X_i * X_j | S_n))$

Then we calculate

$E(X_i^2 | S_n) = (E(X_j^2 | S_n) = Sn/n$

And

$E(X_i*X_j |Sn) = \sum_{i=1}^n\sum_{j=1,i\neq j}^n X_i * X_j = \frac{S_n}{n} * \frac{S_n - 1}{n - 1}$

clds
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